Problem: Solve for $x$ : $3x^2 + 27x + 54 = 0$
Solution: Dividing both sides by $3$ gives: $ x^2 + {9}x + {18} = 0 $ The coefficient on the $x$ term is $9$ and the constant term is $18$ , so we need to find two numbers that add up to $9$ and multiply to $18$ The two numbers $3$ and $6$ satisfy both conditions: $ {3} + {6} = {9} $ $ {3} \times {6} = {18} $ $(x + {3}) (x + {6}) = 0$ Since the following equation is true we know that one or both quantities must equal zero. $(x + 3) (x + 6) = 0$ $x + 3 = 0$ or $x + 6 = 0$ Thus, $x = -3$ and $x = -6$ are the solutions.